Menu Close

If-the-roots-of-ax-2-bx-c-0-are-one-another-s-cube-then-show-that-b-2-2ac-2-ac-a-c-2-

Tinku Tara 0 Comments
Pin




Question Number 205527 by MATHEMATICSAM last updated on 23/Mar/24
If the roots of ax^2 + bx + c = 0 are one another′s cube then show that (b^2 − 2ac)^2 = ac(a + c)^2 .
$$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{are}\:\mathrm{one} \\ $$$$\mathrm{another}'\mathrm{s}\:\mathrm{cube}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left({b}^{\mathrm{2}} \:−\:\mathrm{2}{ac}\right)^{\mathrm{2}} \:=\:{ac}\left({a}\:+\:{c}\right)^{\mathrm{2}} . \\ $$
Answered by A5T last updated on 23/Mar/24
Let the roots be α and β. Then, α=β^3 and β=α^3 α+β=α+α^3 =((−b)/a);αβ=α^4 =(c/a)⇒α=+_− ((c/a))^(1/4) ⇒+_− ((c/a))^(1/4) +_− ((c^3 /a^3 ))^(1/4) =((−b)/a)⇒(√(c/a))+(√(c^3 /a^3 ))+((2c)/a)=(b^2 /a^2 ) ⇒(c/a)+(c^3 /a^3 )+((2c^2 )/a^2 )=(((b^2 −2ac)^2 )/a^4 ) ⇒a^3 c+ac^3 +2a^2 c^2 =ac(a+c)^2 =(b^2 −2ac)^2
$${Let}\:{the}\:{roots}\:{be}\:\alpha\:{and}\:\beta.\:{Then},\:\alpha=\beta^{\mathrm{3}} \:{and}\:\beta=\alpha^{\mathrm{3}} \\ $$$$\alpha+\beta=\alpha+\alpha^{\mathrm{3}} =\frac{−{b}}{{a}};\alpha\beta=\alpha^{\mathrm{4}} =\frac{{c}}{{a}}\Rightarrow\alpha=\underset{−} {+}\sqrt[{\mathrm{4}}]{\frac{{c}}{{a}}} \\ $$$$\Rightarrow\underset{−} {+}\sqrt[{\mathrm{4}}]{\frac{{c}}{{a}}}\underset{−} {+}\sqrt[{\mathrm{4}}]{\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} }}=\frac{−{b}}{{a}}\Rightarrow\sqrt{\frac{{c}}{{a}}}+\sqrt{\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} }}+\frac{\mathrm{2}{c}}{{a}}=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{c}}{{a}}+\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} }+\frac{\mathrm{2}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{\left({b}^{\mathrm{2}} −\mathrm{2}{ac}\right)^{\mathrm{2}} }{{a}^{\mathrm{4}} } \\ $$$$\Rightarrow{a}^{\mathrm{3}} {c}+{ac}^{\mathrm{3}} +\mathrm{2}{a}^{\mathrm{2}} {c}^{\mathrm{2}} ={ac}\left({a}+{c}\right)^{\mathrm{2}} =\left({b}^{\mathrm{2}} −\mathrm{2}{ac}\right)^{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Mar/24
AnOther Way (b^2 − 2ac)^2 = ac(a + c)^2 ⇔(a^2 ((b^2 /a^2 ) − ((2ac)/a^2 )))^2 = ac(a((a/a) + (c/a)))^2 ⇔a^4 ((−(b/a))^2 −2( (c/a)))^2 = a^3 c(1 + (c/a))^2 ⇔((−(b/a))^2 −2( (c/a)))^2 = ((a^3 c)/a^4 )(1 + (c/a))^2 ⇔((−(b/a))^2 −2( (c/a)))^2 = (c/a)(1 + (c/a))^2 ⇔((α+β)^2 −2(αβ))^2 = αβ(1 + αβ)^2 ⇔(α^2 +β^2 )^2 = αβ(1+2αβ + α^2 β^2 ) ⇔α^4 +2α^2 β^2 +β^4 = αβ+2α^2 β^2 + α^3 β^3 α^3 =β ,β^3 =α⇒α^3 β^3 =αβ α^4 =α∙α^3 =αβ , β^4 =β^3 ∙β=αβ ⇔αβ+2α^2 β^2 +αβ = αβ+2α^2 β^2 + αβ ⇔2αβ+2α^2 β^2 =2 αβ+2α^2 β^2 Hence proved
$$\mathrm{AnOther}\:\mathrm{Way} \\ $$$$\left({b}^{\mathrm{2}} \:−\:\mathrm{2}{ac}\right)^{\mathrm{2}} \:=\:{ac}\left({a}\:+\:{c}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\left({a}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{2}{ac}}{{a}^{\mathrm{2}} }\right)\right)^{\mathrm{2}} \:=\:{ac}\left({a}\left(\frac{{a}}{{a}}\:+\:\frac{{c}}{{a}}\right)\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow{a}^{\mathrm{4}} \left(\left(−\frac{{b}}{{a}}\right)^{\mathrm{2}} \:−\mathrm{2}\left(\:\frac{{c}}{{a}}\right)\right)^{\mathrm{2}} \:=\:{a}^{\mathrm{3}} {c}\left(\mathrm{1}\:+\:\frac{{c}}{{a}}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\left(−\frac{{b}}{{a}}\right)^{\mathrm{2}} \:−\mathrm{2}\left(\:\frac{{c}}{{a}}\right)\right)^{\mathrm{2}} \:=\:\frac{{a}^{\mathrm{3}} {c}}{{a}^{\mathrm{4}} }\left(\mathrm{1}\:+\:\frac{{c}}{{a}}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\left(−\frac{{b}}{{a}}\right)^{\mathrm{2}} \:−\mathrm{2}\left(\:\frac{{c}}{{a}}\right)\right)^{\mathrm{2}} \:=\:\frac{{c}}{{a}}\left(\mathrm{1}\:+\:\frac{{c}}{{a}}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\left(\alpha+\beta\right)^{\mathrm{2}} \:−\mathrm{2}\left(\alpha\beta\right)\right)^{\mathrm{2}} \:=\:\alpha\beta\left(\mathrm{1}\:+\:\alpha\beta\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \:\right)^{\mathrm{2}} \:=\:\alpha\beta\left(\mathrm{1}+\mathrm{2}\alpha\beta\:+\:\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \right) \\ $$$$\Leftrightarrow\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\beta^{\mathrm{4}} \:\:=\:\alpha\beta+\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \:+\:\alpha^{\mathrm{3}} \beta^{\mathrm{3}} \\ $$$$\:\:\:\alpha^{\mathrm{3}} =\beta\:,\beta^{\mathrm{3}} =\alpha\Rightarrow\alpha^{\mathrm{3}} \beta^{\mathrm{3}} =\alpha\beta \\ $$$$\:\:\alpha^{\mathrm{4}} =\alpha\centerdot\alpha^{\mathrm{3}} =\alpha\beta\:,\:\beta^{\mathrm{4}} =\beta^{\mathrm{3}} \centerdot\beta=\alpha\beta \\ $$$$\Leftrightarrow\alpha\beta+\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\alpha\beta\:\:=\:\alpha\beta+\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \:+\:\alpha\beta \\ $$$$\Leftrightarrow\mathrm{2}\alpha\beta+\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \:\:=\mathrm{2}\:\alpha\beta+\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \: \\ $$$$\mathcal{H}{ence}\:{proved} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Mar/24
Still another way α+β=α+α^3 =−(b/a)⇒ determinant (((b=−aα^3 −aα))) αβ=αα^3 =α^4 =(c/a)⇒ determinant (((c=aα^4 ))) (b^2 − 2ac)^2 = ac(a + c)^2 ⇒((−aα^3 −aα)^2 − 2a(aα^4 ))^2 = a(aα^4 )(a + aα^4 )^2 ⇒(a^2 (α^3 +α)^2 − 2a^2 α^4 ))^2 = a^2 (a^2 α^4 )(1 + α^4 )^2 ⇒a^4 ((α^6 +2α^4 +α^2 ) − 2α^4 ))^2 = (a^4 α^4 )(1 + α^4 )^2 ⇒a^4 ((α^6 +α^2 ) ))^2 = a^4 α^4 (1 + α^4 )^2 ⇒a^4 (α^(12) +2α^8 +α^4 = a^4 α^4 (1+2α^4 + α^8 ) ⇒a^4 (α^(12) +2α^8 +α^4 )= a^4 (α^4 +2α^8 + α^(12) ) Hence proved
$$\mathrm{Still}\:\mathrm{another}\:\mathrm{way} \\ $$$$\alpha+\beta=\alpha+\alpha^{\mathrm{3}} =−\frac{{b}}{{a}}\Rightarrow\begin{array}{|c|}{{b}=−{a}\alpha^{\mathrm{3}} −{a}\alpha}\\\hline\end{array} \\ $$$$\alpha\beta=\alpha\alpha^{\mathrm{3}} =\alpha^{\mathrm{4}} =\frac{{c}}{{a}}\Rightarrow\begin{array}{|c|}{{c}={a}\alpha^{\mathrm{4}} }\\\hline\end{array} \\ $$$$\left({b}^{\mathrm{2}} \:−\:\mathrm{2}{ac}\right)^{\mathrm{2}} \:=\:{ac}\left({a}\:+\:{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\left(−{a}\alpha^{\mathrm{3}} −{a}\alpha\right)^{\mathrm{2}} \:−\:\mathrm{2}{a}\left({a}\alpha^{\mathrm{4}} \right)\right)^{\mathrm{2}} \:=\:{a}\left({a}\alpha^{\mathrm{4}} \right)\left({a}\:+\:{a}\alpha^{\mathrm{4}} \right)^{\mathrm{2}} \\ $$$$\left.\Rightarrow\left({a}^{\mathrm{2}} \left(\alpha^{\mathrm{3}} +\alpha\right)^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} \alpha^{\mathrm{4}} \right)\right)^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} \alpha^{\mathrm{4}} \right)\left(\mathrm{1}\:+\:\alpha^{\mathrm{4}} \right)^{\mathrm{2}} \\ $$$$\left.\Rightarrow{a}^{\mathrm{4}} \left(\left(\alpha^{\mathrm{6}} +\mathrm{2}\alpha^{\mathrm{4}} +\alpha^{\mathrm{2}} \right)\:−\:\mathrm{2}\alpha^{\mathrm{4}} \right)\right)^{\mathrm{2}} \:=\:\left({a}^{\mathrm{4}} \alpha^{\mathrm{4}} \right)\left(\mathrm{1}\:+\:\alpha^{\mathrm{4}} \right)^{\mathrm{2}} \\ $$$$\left.\Rightarrow{a}^{\mathrm{4}} \left(\left(\alpha^{\mathrm{6}} +\alpha^{\mathrm{2}} \right)\:\right)\right)^{\mathrm{2}} \:=\:{a}^{\mathrm{4}} \alpha^{\mathrm{4}} \left(\mathrm{1}\:+\:\alpha^{\mathrm{4}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{4}} \left(\alpha^{\mathrm{12}} +\mathrm{2}\alpha^{\mathrm{8}} +\alpha^{\mathrm{4}} =\:{a}^{\mathrm{4}} \alpha^{\mathrm{4}} \left(\mathrm{1}+\mathrm{2}\alpha^{\mathrm{4}} \:+\:\alpha^{\mathrm{8}} \right)\right. \\ $$$$\Rightarrow{a}^{\mathrm{4}} \left(\alpha^{\mathrm{12}} +\mathrm{2}\alpha^{\mathrm{8}} +\alpha^{\mathrm{4}} \right)=\:{a}^{\mathrm{4}} \left(\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{8}} \:+\:\alpha^{\mathrm{12}} \right) \\ $$$$\mathcal{H}{ence}\:{proved} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *